Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 12 - Exercises and Problems - Page 216: 34



Work Step by Step

We know that in the given scenario $\sum T_p=0$ $\implies Tsin\theta\times 2.3=(M+m)g\times \frac{2.3}{2}$ This simplifies to: $T=\frac{(M+m)g}{2sin\theta}$ We plug in the known values to obtain: $T=\frac{(66+8.2)\times 9.8}{2sin\theta}$ $T=\frac{363.6}{sin\theta}$ We also know that $tan\theta=\frac{h}{2.3}$ , hence $sin\theta=\frac{h}{\sqrt{h^2+2.3^2}}$ As given that the maximum tension is 800N so $T<800N$ $\implies 363.6\sqrt{h^2+(2.3)^2}<800h$ After taking square on both sides, we obtain: $(363.6)^2(h^2+2.3^2)\frac{363.6\times 2.3}{\sqrt{(800)^2-(363.6)^2}}=1.2m$ Thus, the minimum height should be greater than 1.2m
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